Problem: Evaluate $\int\sin^2\Big(\dfrac14x\Big)\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac12x-2\sin\Big(\dfrac12x\Big)+C$ (Choice B) B $\dfrac12x-4\sin\Big(\dfrac14x\Big)+C$ (Choice C) C $\dfrac12x-\sin\Big(\dfrac12x\Big)+C$ (Choice D) D $ x-2\sin\Big(\dfrac12x\Big)+C$
Answer: There are no odd powers of either $~\sin~$ or $~\cos\,$, so we use the double angle formula for $~\cos~$ that relates to $~\sin^2$. We know that $~~~\cos2t=1-2\sin^2t\,$. Therefore, $ \sin^2t=\dfrac{1-\cos2t}2$ Then $ \sin^2\Big(\dfrac14x\Big) =\dfrac{1-\cos\big(\frac12x\big)}2=\dfrac12\Big(1-\cos\big(\frac12x\big)\Big) $ Now we can integrate. $ \int\sin^2\Big(\dfrac14x\Big)\,dx=\dfrac12\int\Big(1-\cos\big(\frac12x\big)\Big)\,dx$ $ ~~~=\dfrac12x-\sin\Big(\dfrac12x\Big)+C$